Elitmus
Exam
Numerical Ability
Number System
Let R1 and R2 respectively denote the maximum and minimum possible remainders when (276)n is divided by 91 for any natural number n,n>= 144. Find R1+R2.
Read Solution (Total 10)
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- But the least value for n can be= 144 (as given in question, n>=144)
So, 3n/91= 3*144/91= 68
now, the highest remainder can be when it reaches 68+3+3+3+3+3+3+3= 89
(as each time we increment n by 1 the remainder increases by 3,
3*145/91= 71- remiander
3*146/91= 74
3*147/91=77
... so on and so far upto 89 after which it becomes 89+3= 92 which means it crosses 91)
Hence least remainder would be 92 -91=1
and highest remainder = 89
R1+ R2= 89+1 = 90
- 8 years agoHelpfull: Yes(20) No(4)
- The remainder when 276 divided by 91 is 3 hence
remainder of (276)n/91 = remainder 3n/91
Taking n = 1,2,3,4,5
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats
Hence R1 = 81 and R2 = 1
R1+R2 = 81+1 = 82 - 8 years agoHelpfull: Yes(5) No(6)
- The remainder when 276 divided by 91 is 3 hence
remainder of (276)n/91 = remainder 3n/91
Taking n = 1,2,3,4,5
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats
Hence R1 = 81 and R2 = 1
R1+R2 = 81+1 = 82 - 8 years agoHelpfull: Yes(2) No(0)
- any number say m when divide another say n . contain minimum remainder 0 and maximum m-1.
so in case of (276)n / 91 . no need to calculate answer will be R1+R2= 0+(91-1)= 90 - 8 years agoHelpfull: Yes(1) No(1)
- @himanshu send me study material for preparing elitmus n my email id- adikumar1995@gmail.com
- 8 years agoHelpfull: Yes(0) No(0)
- ans would be 91..the max remainder can be 90n the minimum remainder can be 1 so R1+R2 would be.(90+1=91)
- 8 years agoHelpfull: Yes(0) No(0)
- sorry in my earlier solution i did a mistake ...276*n/91 the remainder will be 3*n/91,now if i devide n by 91 the maximum remainde can be 90 so 3*90=270 gain devide it by 91 the remainder would be 88..
now in the menimum case n/91 the minimum remainder can be 1 so3*1/91 rem is 3 hence r1=3,r2=87 so total sum would be=87+3=91
- 8 years agoHelpfull: Yes(0) No(2)
- he remainder when 276 divided by 91 is 3 hence
remainder of (276)n/91 = remainder 3n/91
Taking n = 1,2,3,4,5
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats
Hence R1 = 81 and R2 = 1
R1+R2 = 81+1 = 82 - 7 years agoHelpfull: Yes(0) No(0)
- The remainder when 276 divided by 91 is 3 hence
remainder of (276)n/91 = remainder 3n/91
Taking n = 1,2,3,4,5
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats
Hence R1 = 81 and R2 = 1
R1+R2 = 81+1 = 82 - 7 years agoHelpfull: Yes(0) No(0)
- The remainder when 276 divided by 91 is 3 hence,
remainder of (276)n/ 91 = remainder 3n/91
Taking n = 1,2,3,4,5.
We get the possible remainders as 3, 9, 27, 81, 61, 1 respectively.
For n>5, the pattern repeats.
Hence, R1 = 81 and R2 = 1.
and, R1 + R2 = 81 + 1 = 82. - 5 years agoHelpfull: Yes(0) No(0)
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